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Amazon Redshift COMPOUND SORTKEY - does insertion order matter?


By : Derek Randall Darr
Date : September 16 2020, 01:00 AM
should help you out Assuming the same behaviour for INSERT INTO as for loading via the COPY command, there is no need to order the records first. According to the AWS docs all the following constraints be fulfilled in order to add the records to sorted region of the table - in your example you have a COMPOUND SORTKEY of 2 columns:
The table uses a compound sort key with only one sort column. The sort column is NOT NULL. The table is 100 percent sorted or empty. All the new rows are higher in sort order than the existing rows, including rows marked for deletion. In this instance, Amazon Redshift uses the first eight bytes of the sort key to determine sort order.
code :


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Redshift: Should the sortkey contain the distkey?


By : Mart
Date : March 29 2020, 07:55 AM
wish helps you Dist key does not affect the order in which rows are stored in each node/slice/block. Sort key (or natural order in the absence of such) defines the order.
If you expect frequent queries with company_id and you want to achieve maximum performance, make company_id the main sort key (COMPOUND or default, not just INTERLEAVED).

Redshift DISTKEY / SORTKEY


By : yes
Date : March 29 2020, 07:55 AM
I hope this helps you . The DISTKEY distributes rows amongst slices.
In your example, all rows with a given orderdate would be located in the same slice. That means that all columns for those rows are in that slice.

Redshift POSIX regex order does not matter


By : Melcamp
Date : March 29 2020, 07:55 AM
I wish did fix the issue. This is probably most clearly written as ANDed separate regular expression matches. To ensure that you don't match e.g. pineapple when looking for apple, you need to check that on either side of the search term is either a space character or the beginning/end of the line:
code :
SELECT full_term FROM data_table
WHERE full_term ~ '(^|\\s)apple(\\s|$)'
  AND full_term ~ '(^|\\s)juice(\\s|$)'

Why does the Order of GROUPBY and ORDERBY matter in Redshift?


By : jenny
Date : March 29 2020, 07:55 AM
around this issue GROUP BY implicitly requires the data to be sorted. If your ORDER BY is different the data will have to be sorted twice.
And, since Redshift distributes it's data among multiple nodes, that may mean the data has to be sent to all nodes twice.

MongoDB Compound Indexes - Does the sort order matter?


By : Herman Johannes
Date : March 29 2020, 07:55 AM
help you fix your problem You can think of MongoDB single-field index as an array, with pointers to document locations. For example, if you have a collection with (note that the sequence is deliberately out-of-order):
code :
[collection]
1: {a:3, b:2}
2: {a:1, b:2}
3: {a:2, b:1}
4: {a:1, b:1}
5: {a:2, b:2}
db.collection.createIndex({a:1})
[index a:1]
1: {a:1} --> 2, 4
2: {a:2} --> 3, 5
3: {a:3} --> 1
db.collection.find().sort({a:1})
db.collection.find().sort({a:-1})
db.collection.createIndex({a:1, b:1})
[index a:1, b:1]
1: {a:1, b:1} --> 4
2: {a:1, b:2} --> 2
3: {a:2, b:1} --> 3
4: {a:2, b:2} --> 5
5: {a:3, b:2} --> 1
db.collection.find({a:2}).sort({b:1})
db.collection.find().sort({b:1})
db.collection.find({b:1})
db.collection.find({a:2}).sort({b:-1})
a | b
--|--
1 | 1
1 | 2
2 | 1
2 | 2
2 | 3
3 | 1
3 | 2
a | b
--|--
1 | 1
1 | 2
2 | 1 <-- walk this block forward to satisfy
2 | 2 <-- find({a:2}).sort({b:1})
2 | 3 <--
3 | 1
3 | 2
a | b
--|--
1 | 1
1 | 2
2 | 1  <-- walk this block backward to satisfy
2 | 2  <-- find({a:2}).sort({b:-1})
2 | 3  <--
3 | 1
3 | 2
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